"! # $
Solutions to Odd-numbered Problems in:
The Calculus,
Algebraically
19.
21.
Limits, Informally
23.
Answers to exercises
from 4.1, page 158
25.
1.
3.
5.
7.
9.
11.
13.
15.
17.
lim x = 3
x→3
27.
lim x2 = 1
x→1
lim
x→0
1
= undefined
x
29.
1
1
lim
=
x→2 x
2
lim
x→∞
x+1
=1
x−1
lim x sin(x) = 0
x→π
lim
cos(x)
= cos(1)
x
lim
sin2 (x)
= sin2 (1)
x
x→1
x→1
x3 + 5x2 − 8x − 48
= 49
x→3
x−3
Again, factor the top into (x + 4)2 (x − 3).
lim
2x3 + x − 6x2 − 2
= 19
x−3
Factor 2x3 + x − 6x2 − 2 into (2x2 + 1)(x − 3).
lim
x→3
lim x2 − 4 = −3
x→1
lim cos(2x) = −1
x→ π
2
lim
x→−1
lim
x→1
x+1
=1
x+1
35.
lim
x−1
=0
x+1
∆x→0
=
x2 − 1
lim
=2
x→1 x − 1
The trick to this one is factoring
x2 − 1 into (x + 1)(x − 1).
Then the (x − 1) in the numerator
and denominator cancel and f (x) = x + 1.
=
=
=
527
lim
∆x→0
lim
∆x→0
lim
∆x→0
3
x
2
3
(x
4
3 2
x
4
+ ∆x)2 − 2 − ( 34 x2 − 2)
∆x
+ 23 x∆x + 34 ∆x2 − 2 − 34 x2 + 2
∆x( 23 x + 43 ∆x)
∆x
3
3
x + ∆x
2
4
∆x
528
Answers, Notes, and Hints
37.
lim
(x + ∆x) −
lim
x + ∆x −
∆x→0
=
=
=
∆x→0
lim 1 −
∆x→0
1 − x2
(x+∆x)2
4
x2
4
∆x
2x
−
4
4
− (x −
∆x
− 2x∆x
−
4
∆x2
4
∆x
x2
)
4
−x+
x2
4
39.
lim
∆x→0
=
=
=
=
lim
5
(x+∆x)+1
lim
5
− ( x+1
)
∆x
5(x+1)−5(x+∆x+1)
(x+∆x+1)(x+1)
∆x
∆x→0
−5∆x
x2 +2x+1+x∆x+∆x
∆x
∆x→0
lim −
∆x→0
5
− (x+1)
2
x2
5
+ 2x + 1 + x∆x + ∆x
41.
lim
∆x→0
=
=
=
lim
(x+∆x)
(x+∆x)2 +2
−
x
x2 +2
∆x
(x+∆x)(x2 +2)−x(x2 +2x∆x+∆x2 +2)
(x2 +2)((x+∆x)2 +2)
∆x
+ 2 − x∆x)
lim
∆x→0 ∆x(x2 + 2)((x + ∆x)2 + 2)
∆x→0
xs
∆x(−x2
2−x2
(x2 +2)2
43.
lim
∆x→0
=
=
=
=
lim
1
2−2(x+∆x)
−
1
2−2x
∆x
(2−2x)−(2−2x−2∆x)
(2−2x−2∆x)(2−2x)
∆x→0
∆x
2∆x
4∆x(1 − 2x + 2x2 − ∆x + x∆x)
1
lim
∆x→0 2 − 4x + 4x2 − 2∆x + 2x∆x
lim
∆x→0
1
2(1−x)2
This is an example of a function that is not defined at
x = 0, but whose limit exists, namely, when x → 0,
y → .5.
45.
47.
lim |x| = 0
x→0
lim
x→1
√
x−1
= 0.5
x−1
Lines, Powers, and
Polynomials
Answers to exercises
from 4.2, page 173
1.
f 0 (x) = 5
3.
f 0 (x) = 0, since the linear formula here is
f (x) = mx + 6 and m = 0.
5.
f 0 (x) = 89
7.
f 0 (x) = 1
9.
f 0 (x) = 44
11.
f 0 (x) = −1
13.
f 0 (x) = 2x
15.
f 0 (x) = 17x16
17.
f 0 (x) = 4
19.
f 0 (x) = −2x
21.
f 0 (x) = 0
23.
f 0 (x) = 0
#
Neither Newton nor Leibniz married. Newton devoted his
life to his work. In his words, “. . . in order to catch at
a shadow, I had sacrificed my peace, a truly substantial
thing.”
"
25.
f 0 (x) = 2 − 30x
27.
f 0 (x) = 156x12
29.
f 0 (x) = 297x2 − 45
31.
f 0 (x) = 2x + 15x2
33.
f 0 (x) = 28x3 − 42x
35.
f 0 (x) = 4
37.
f 0 (x) = 15x4 − 16x3 + 10x − 6
41. a. In the year 1992 (t = 3), the number of visitors with
RV’s entering the National Park System was decreasing by approximately 25,000 people per year. We
know this because R0 (t) = −25.
!
Appendix E
b. Refering to the explanation given in answer a., this
constant derivative tells us that the number of visitors with RV’s is decreasing at this rate every year
from 1979 through 1998.
43.
b0 (t) = .1 Therefore, the rate of growth from 1990 to
1998 was .1 million dollars per year, which is constant
for each year.
45. a. K 0 = t
b. H 0 =
3
t
5
'
Newton’s consuming passion for understanding gave us
much, but at great personal cost. Thomas Huxley, in 1932,
said “If we evolved a race of Isaac Newtons, that would not
be progress. For the price Newton had to pay for being a
supreme intellect was that he was incapable of friendship,
love, fatherhood, and many other desirable things. As a
man he was a failure; as a monster he was superb.”
529
53.
The derivative of a cubic function is a quadratic function. If a cubic function had more than 3 roots, then
it’s derivative would have more than 2 roots. But a
quadratic function can’t have more than 2 roots by the
previous problem. So a cubic function can’t have more
than 3 roots.
55. a. The limit can be written as the limit of a
∆g
difference quotient.
Therefore,
lim
=
∆x→0 ∆x
g(a + ∆x) − g(a)
lim
∆x→0
∆x
b. Since g(x) = cf (x), their difference quotients are
equivalent.
$
c. By factoring out the limit of the constant, c, we have
a product of two limits.
d. Since,
the constant c is not affected as ∆x →
0, lim c = c.
∆x→0
e. Since we were given that cf is a function differentiable
at x = a, the limit exists as ∆x → 0, and we obtain
f 0 (a).
The proof of Proposition 4.9.
∆(f − g)
lim
&
%
∆x→0
∆x
(f (a + ∆x) − g(a + ∆x)) − (f (a) − g(a))
a. = lim
∆x→0 ∆x
f (a + ∆x) − f (a)
g(a + ∆x) − g(a)
47. a. The ball’s velocity equals B 0 (t) = −1.6t + 2.
b. = lim
−
∆x→0
∆x
∆x
b. The baseball reached its maximum height of 1.25 meg(a + ∆x) − g(a)
f (a + ∆x) − f (a)
ters at a time of 1.25 seconds.
− lim
c. = lim
∆x→0
∆x→0
∆x
∆x
c. The ball’s acceleration equals B 00 (t) = −1.6.
d. = f 0 (a) − g 0 (a)
49. a. The ball’s velocity equals R0 (t) = −1.6t + 50.
b. The baseball reached its maximum height of 781.25
meters at a time of 31.25 seconds.
c. The ball’s acceleration equals R00 (t) = −1.6.
d. The height of a object thrown form the surface of the
moon at time t with initial velocity v can be modeled
by the equation H(t) = −.8t2 − vt.
The Product Rule
e. Its velocity equals H 0 (t) = −1.6t − v.
Answers to exercises
from 4.3, page 183
f. Its acceleration equals H 00 (t) = −1.6.
g. The acceleration is constant, no matter what the object. This number, −1.6 meters per second per second, is the acceleration due to gravity on the moon.
Galileo was the first to find out that all objects falling
1.
f 0 (x) = 2x(sin(x)) + x2 (cos(x))
to Earth have a constant acceleration of 9.8 meters
= 2x sin(x) + x2 cos(x)
per second per second, regardless of their mass. Since
the moon is considerably less massive than the Earth,
its gravitational pull is less forceful, and objects fall
3.
f 0 (x) = 0(x4 + 7) + 12 (4x3 )
to its surface less quickly.
= 2x3
51.
Between any two roots of f there must be a peak or a
valley in the graph of f , so the number of peaks and
valleys is at least n − 1. At a peak or valley, the slope is
zero, that is, f 0 is zero. Therefore the number of roots
of f 0 is at least n − 1.
57.
Many of these, such as this one, can be done either by the
product rule or the constant rule! For this set of exercises,
use the product rule.
530
Answers, Notes, and Hints
5.
f 0 (x)
=
=
(9x2 )(2x − 14 ) + (3x3 + 6)2
24x3 − 49 x2 + 12
7.
f 0 (x)
=
=
x(5x2 − 4) +
10x3 − 4x
f 0 (x)
27.
=
x2
(10x)
2
29.
f 0 (x)
31.
f 0 (x)
The independent variable has been announced as x (via
“f (x)”) so all other letters are constant.
9.
f 0 (x)
=
=
2 sin(x) + 2x cos(x) + sin(x)
2x cos(x) + 3 sin(x)
11.
f 0 (x)
=
=
8(15x2 ) + (8x − 3)(30x)
360x2 − 90x
'
If you’re having trouble mastering the Multiplication Rule,
take some comfort from the fact that Leibniz puzzled over
it himself for an entire year, not quite believing, it is said,
that (f g)0 wasn’t f 0 g 0 . Of course, his position was somewhat different from yours. He was inventing the calculus.
He probably thought that if f 0 g 0 didn’t work, then perhaps
he had defined “derivative” incorrectly! Even the greatest
geniuses have doubts.
&
13.
=
=
=
−24x2 (5x4 + 20x2 )
+(10 − 8x3 )(20x3 + 40x)
−280x6 − 800x4 + 200x3 + 400x
(1 + 15x2 )(x + 2 − x3 )
+(x + 2 + 5x3 )(1 − 3x2 )
4 + 2x + 24x2
−16x3 − 30x5
35.
f 0 (x)
=
(2ax + b − ax2 − bx − c) cos(x)
−(2ax + b + ax2 + bx + c) sin(x)
$
f 0 (x)
=
81x2 sin(x)(− sin(x))
+ cos(x)(81x2 cos(x) + sin(x)(162x))
−81x2 sin2 (x) + 81x2 cos2 (x)
+162x sin(x) cos(x)
39.
f 0 (x)
37.
=
=
=
(5 sin2 (x))
+(5x + 2)(2 cos(x) sin(x))
5 sin2 (x) + 5x cos(x) sin(x) + 4 cos(x) sin(x)
'
%the age of fifty; but the person he had in mind
$
asked for time to reflect. This gave Leibniz
time to reflect, too, and so he never married.
f 0 (x) = 11
ab
4
Since a and b are constants, you don’t need to use the
product rule on them.
15.
=
− sin2 (x)
17.
f 0 (x)
=
=
1(12x2 + 5x9 ) + (x + 3)(24x + 45x8 )
36x2 + 50x9 + 72x + 135x8
19.
f 0 (x)
=
=
100 cos(x) + (100x − 7)(− sin(x))
100 cos(x) − 100x sin(x) + 7 sin(x)
21.
f 0 (u)
=
=
(3u2 − 5)(3u − 4) + (u3 − 5u)3
12u3 − 12u2 − 30u + 20
23.
f 0 (x)
=
(3x2 )0 (20
=
=
f 0 (x)
=
cos(x)(15x12 + 2x3 − x − 33)
+ sin(x)(180x11 + 6x2 − 1)
15x12 cos(x) + 2x3 cos(x) − x cos(x)
−33 cos(x) + 180x11 sin(x) + 6x2 sin(x) − sin(x)
Leibniz never married; he had considered it at
f 0 (x)
25.
=
=
=
+
cos2 (x)
&
41.
f 0 (x)
=
=
43.
f 0 (x)
=
(3x2 )
− sin(x)) +
((20)0 − (sin(x))0 )
(6x)(20 − sin(x))
+(3x2 )((0) − (cos(x)))
120x − 6x sin(x) − 3x2 cos(x)
(−1 − 9x2 )(7 + x5 )+
(2 − x − 3x3 )(5x4 )
−24x7 − 6x5 + 10x4
−63x2 − 7
– Bernard Le Bovier Forenelle
=
%
(− sin(x) − 4)(4x + 12)(−14x − sin(x))
+4(cos(x) − 4x)(−14x − sin(x))+
(−14 − cos(x))(4x + 12)(cos(x) − 4x)
56x2 sin(x) + 16x2 cos(x) + 200x sin(x)
−64 cos(x) + 4x sin2 (x) − 4x cos2 (x) + 12 sin2 (x) − 12 cos2 (x)+
48 sin(x) − 168 cos(x) + 682 cos(x) + 682x2
+1344x − 4 cos(x) sin(x)
(3x2 + 8.4x)(3x + 3)(sin(x) + 15x2 )(−4x9 )
+(x3 + 4.2x2 )(3)(sin(x) + 15x2 )(−4x9 )+
(x3 + 4.2x2 )(3x + 3)(cos(x) + 30x)(−4x9 )
+(x3 + 4.2x2 )(3x + 3)(sin(x) + 15x2 )(−36x8 )
153x12 sin(x) − 1019.6x11 sin(x) − 894.6x10 sin(x)
−12x13 cos(x) − 516x12 cos(x)−
504x11 cos(x) − 2160x14 − 10476x13 − 8316x12
47. a. The Product Rule
b. Combining like terms
Appendix E
531
c. Since f (x) =
√
√ √ 0
x, (f (x) · f (x))0 =
x · x = x0 .
9.
F 0 (r)
=
=
d. Combining the equalities from b and c above,
2f 0 (x)x.5 = 1, and solving algebraically for f 0 (x),
we get 12 x1.5 .
11.
e. Yes
f 0 (x)
2
3f 0 (x)x 3
b. (f (x) · f (x) ·
=
c. (f (x) · f (x) ·
f (x))0
= (x1 )0 = 1
d. f 0 (x) =
=
=
49. a. (f (x) · f (x) · f (x))0 = (f 0 (x) · f (x) · f (x)) +
(f (x) · f 0 (x) · f (x)) + (f (x) · f (x) · f (x)0 )
f (x))0
=
=
13.
f 0 (x)
=
=
=
1 −2
x 3
3
15.
f 0 (x)
=
=
=
=
The Quotient Rule
Answers to exercises
from 4.4, page 191
1.
f 0 (x)
=
=
=
3.
f 0 (x)
=
(5x)0 (12)−(120 )(5x)
(12)2
(5)(12)−(0)(5x)
144
5
12
(7x+6)0 (x3 )−(x3 )0 (7x+6)
(x3 )2
(7+0)(x3 )−(3x2 )(7x+6)
x6
− 14x+18
4
x
−3
−4
=
=
= −14x
5.
f 0 (x)
=
=
=
7.
f 0 (x)
=
=
=
− 18x
(8x)0 (x−15)−(x−15)0 (8x)
(x−15)2
(8)(x−15)−(1−0)(8x)
2
x −30x+225
−120
x2 −30x+225
0
0
(sin(x)) (x)−(x) (sin(x))
(x)2
(cos(x))(x)−(1)(sin(x))
x2
x cos(x)−sin(x)
x2
17.
f 0 (x)
=
=
=
=
19.
f 0 (x)
(GM m)0 (r 2 )−(r 2 )0 (GM m)
(r 2 )2
(0)(r 2 )−(2r)(GM m)
r4
m
− 2GM
r3
( 43 x−4 )0 (cos(x))−(cos(x))0 ( 43 x−4 )
cos2 x
(−3x−5 cos(x))+( 34 x−4 sin(x))
cos2 x
x−1 cos(x)− 41 sin(x)
)
−3x−4 (
cos2 x
(x−3 +2)0 (2x−1)−(2x−1)0 (x−3 +2)
(2x−1)2
−3x−4 (2x−1)−2(x−3 +2)
(2x−1)2
3x−4 −8x−3 −4
4x2 −4x+1
(3+x)0 (x−3 )−(x−3 )0 (3+x)
(x−3 )2
(1)(x−3 )−(−3x−4 )(3+x)
(x−3 )2
x−3 +9x−4 +3x−3
x−6
9x−1 +4
9
x
(sin(x))0 (x−2 )−(x−2 )0 (sin(x))
(x−2 )2
(cos(x))(x−2 )−(−2x−3 )(sin(x))
x−4
x−2 cos(x)+2x−3 sin(x)
x−4
cos(x)+2x−1 sin(x)
x−2
0
0
(1)
= (1) (x)−(x)
x2
1
= − x2
Using the trigonometric identity, sin2 x +
cos2 x = 1, we simplified so that f (x) = x1 .
21.
f 0 (x)
=
=
=
23.
f 0 (x)
=
=
=
(2x3 −6x2 +x−2)0 (x−3)−(x−3)0 (2x3 −6x2 +x−2)
(x−3)2
(6x2 −12x+1)(x−3)−(1)(2x3 −6x2 +x−2)
x2 −6x+9
−8x3 +24x2 −1
2
x −6x+9
[2x(4x+12)]0 (5+2x)−(5+2x)0 (2x(4x+12))
(5+2x)2
[(2)(4x+12)+(2x)(4)](5+2x)−(2x(4x+12))(2)
25+20x+4x2
16x2 +80x+120
4x2 +20x+25
532
24.
Answers, Notes, and Hints
f 0 (x)
=
=
=
26.
f 0 (x)
=
=
=
(3x2 +a cos(x))0 (2a+x)−(2a+x)0 (3x2 +a cos(x))
(2a+x)2
(6x−a sin(x))(2a+x)−(1)(3x2 +a cos(x))
4a2 +4x+x2
12ax+3x2 −2a2 sin(x)−ax sin(x)−acosx
4a2 +4ax+x2
[(2x2 −5x+1)(x−1)]0 (x+1)−(x+1)0 ((2x2 −5x+12)(x−1))
(x+1)2
[(2x2 −5x+1)(1)+(4x−5)(x−1)](x+1)−(1)((2x2 −5x+12)(x−1))
(x+1)2
3x3 −x2 −24x+19
(x+1)2
'
It is rare to find learned men who are clean,
do not stink, and have a sense of humor.
– said about Leibniz, either by Charles Louis de
Secondat Montesquieu or the Duchess of Orléans
&
28.
f 0 (x)
=
=
=
34.
f 0 (x)
$
%
(2(x2 −5))0 (x2 +x)−(x2 +x)0 (2(x2 −5))
(x2 +x)2
4x(x2 +x)−(2x+1)(2x2 −10)
(x2 +x)2
2x2 +20x+10
(x2 +x)2
]0 (5x−3 ) + (5x−3 )0 (56 sin(x)−4x)(12x+6)
= [ (56 sin(x)−4x)(12x+6)
(x+12)
x+12
0
0
((56 sin(x)−4x)(12x+6))
= [ [(56 cos(x)−4x)(12x+6)] (x+12)−(x+12)
] + (−15x−4 ) (56 sin(x)−4x)(12x+6)
(x+12)2
x+12
=
=
36.
38.
40.
[(56 sin(x)−4x)(12)+(12x+6)(56 cos(x)−4)](x+12)−(12)(56 sin(x)−4x)(12x+6)
(56 sin(x)−4x)(12x+6)
+ ( −15
)
x2 +24x+144
x4 )(
x+12
2
2
−7056x sin(x)−4032 sin(x)+672x cos(x)+8064x cos(x)+4032 cos(x)+480x −888x−288
x2 +24x+144
sin(x)−720x2 −360x
− 10080x sin(x)+5040
x5 +12x4
= (17x − 4x2 + x3 )0 = 17 − 8x + 3x2
0
0
0
(sin(x))
sin(x)
= (sin(x)) (cos(x))−(cos(x))
(tan(x))0 = cos(x)
cos2 x
f 0 (x)
(cot(x))0 =
cos(x)
sin(x)
0
=
=
=
(cos(x))(cos(x))−(− sin(x))(sin(x))
cos2 x
cos2 x+sin2 x
2
cos x
1
2
cos2 x = sec x
=
(cos(x))0 (sin(x))−(sin(x))0 (cos(x))
sin2 x
(− sin(x))(sin(x))−(cos(x))(cos(x))
sin2 x
− sin2 x−cos2 x
2
sin x
− sin12 x = csc2 x
=
=
=
Appendix E
533
17.
(sec(5z))0
19.
√
3
( x2 ) 0
21.
((1 + (1 + x)2 )2 )0
The Chain Rule
Answers to exercises
from 4.5, page 197
1.
((x − 5)7 )0
= 7(x − 5)6 (x − 5)0
= 7(x − 5)6
3.
(sin100 w)0
5.
(sin(2y − 6))0
7.
((cos(x))53 )0
9.
((2w − 6).4 )0
11.
(sec2 (x2 ))0
= 100 sin99 w(sin w)0
= 100 sin99 w(cos w)
= cos(2y − 6)(2y − 6)0
= 2 cos(2y − 6)
= 53(cos(x))52 (cos(x))0
= −53(cos(x))52 (sin(x))
= .4(2w − 6)−.6 (2w − 6)0
= .8(2w − 6)−.6
√
( t − 1)0
=
=
1
− 12
(t
2 (t − 1)
√1
2 t−1
15.
√
( 3 sin y)0
=
=
=
1
3
1
3
3
23.
(sin(x − sin(x − 1)))0
25.
(sin(3x2 + 2x))0
27.
(cos(sin2 (x)))0
29.
(sin(x2 + (13 − x))4 )0
$
dz dy
dz
=
.
dy dx
dx
You could almost imagine that the notation is trying to tell
us something. Heinrich Hertz once wrote:
2
(sin y)
= 2(1 + (1 + x)2 )(1 + (1 + x)2 )0 (1 + x)0
= 2(1 + (1 + x)2 )(0 + 2(1 + x))(1)
= (4 + 4x)(1 + (1 + x)2 )
The Chain Rule looks very natural in Leibniz’s notation:
− 1)0
sin y − 3 (sin y)0
2
sin y − 3 (cos y)
cos y
√
3
2
2
= ( 31 (x2 )− 3 )(x2 )0
2
= ( 13 (x2 )− 3 )(2x)
= 23 x−4/3 x
= 32 x−1/3
'
= 2 sec(x2 ) sec(x2 ) tan(x2 )(x2 )0
= 4x sec2 (x2 ) tan(x2 )
13.
= sec(5z) tan(5z)(5z)0
= 5 sec(5z) tan(5z)
&
One cannot escape the feeling that
these mathematical formulae have an
independent existence and an intelligence of their own, that they are wiser
than we are, wiser even than their discovers, that we get more out of them
than was originally put into them.
= cos(x − sin(x − 1))(x − sin(x − 1))0 (x − 1)0
= cos(x − sin(x − 1))(1 − cos(x − 1))(1)
= (cos(x2 + 2x))(x2 + 2x)0
= cos(x2 + 2x)(2x + 2)
= (− sin(sin2 (x)))((sin(x))2 )0 (sin(x))0
= − sin(sin2 (x))(2 sin(x))(cos(x))
= −2 sin4 (x) cos(x)
= (4 sin(x2 (13 − x))3 )(sin(x2 (13 − x)))0 (x2 (13 − x))0
= (4 sin(x2 (13 − x))3 )(cos(x2 (13 − x)))(2x − 1)
%
534
Answers, Notes, and Hints
31.
(sin2 (cos(x)))0
33.
((cos(6x − 1)) 4 )0
= (2 sin(cos(x)))(sin(cos(x)))0 (cos(x))0
= (2 sin(cos(x)))(cos(cos(x)))(− sin(x))
3
1
= ( 43 (cos(6x − 1))− 4 )(cos(6x − 1))0 (6x − 1)0
1
= ( 43 (cos(6x − 1))− 4 )(− sin(6x − 1))(6)
sin(6x−1)
= − 9√
4
2
cos(6x−1)
√ 0
2
37. Start by computing ( x)
in two different
√ 0
2
ways. First, ( x)
= (x)0 = 1. On the
√ 0
2
other hand, using the Chain Rule, ( x)
=
√ √ 0
2 x ( x) . Putting these two expressions for
√ 0
√ √ 0
2
together, we get 2 x ( x) = 1,
( x)
√ 0
which we solve algebraically to obtain ( x) =
1
√ .
2 x
11
7
7
11
39. g(f (x)) = (x ) , which equals x . Then,
(g(f (x)))0 = 11x10 . We know from the Chain
11
Rule, that (g(f (x)))0 also equals 7(x 7 )6 (f 0 (x)).
Combining these two equalities, we get 11x10 =
11
7(x 7 )6 (f 0 (x)). Then, algebraically, we solve:
10
11 47
0
(10− 66
7 ) =
f 0 (x) = 11x66 = 11
7 x
7 x = f (x).
7x
(sin(arcsin(x)))0
cos(arcsin(x))(arcsin(x))0
41.
q
1 − sin2 (arcsin(x)) (arcsin(x))0
√
1 − x2 (arcsin(x))0
(arcsin(x))0
43.
Answers to exercises
from 4.6, page 202
Step
Step
Step
Step
Step
A. This is a subtraction, 3x minus 5
B. f 0 (x) = (3x)0 − (5)0
C. 3x and 5 are basic functions.
D. f 0 (x) = 3.
F. (optional) Diagram the function.
= x0
= 1
= 1
= 1
1
= √1−x
2
(sec(arcsec(x)))0
sec(arcsec(x))
tan(arcsec(x))(arcsec(x))0
p
0
2
x sec (arcsec(x))
√ − 1 (arcsec(x))0
2
x x − 1 (arcsec(x))
(arcsec(x))0
Differentiating
Complicated Functions
1.
7
=
=
=
=
=
x0
1
1
1
√1
x x2 −1
Appendix E
535
3
3.
x
−
5
Step A. This is a composite function. The outside is √ , and the inside is a quotient,
−1/2
z
z
Step B. f 0 (z) = ( 12 z−3
( z−3
)0
Step C. The quotient has top z and bottom z − 3.
−1/2
−1/2
(z−3)(z)0 −z(z−3)0
−3
z
z
Step D. f 0 (x) = 21 z−3
= 21 z−3
(z−3)2
(z−3)2
Step F. (optional) Diagram the function.
z
z−3
5.
Step
Step
Step
Step
Step
Step
A. This is a composite function. The outside is ( )2 , and the inside is sin(t3 + 4).
B. f 0 (t) = 2 sin(t3 + 4)(sin(t3 + 4))0
C. This is a composite function. The outside is sin( ) and the inside is t3 + 4.
D. f 0 (t) = 2 sin(t3 + 4) cos(t3 + 4)(t3 + 4)0
E. f 0 (t) = 2 sin(t3 + 4) cos(t3 + 4)(3t2 )
F. (optional) Diagram the function.
2
sin
7.
Step
Step
Step
Step
t3 + 4
√
√
A. This is a √
subtraction,
x minus 3x
√
B. f 0 (x) = ( x)0 − ( 3x)0
C. There is a composite function, where the outside is √
1
D. f 0 (x) = 2√
− √1 (3x)0
x
2
3x
1
Step E. f 0 (x) = 2√
− √3
x
2 3x
Step F. (optional) Diagram the function.
and the inside is 3x.
z
.
z−3
536
Answers, Notes, and Hints
√
x
−
√
3x
In an earlier bubble, we quoted Hertz wondering if formulas
were somehow smarter than the humans who wrote them.
Euler, apparently, had a nagging feeling that his pencil was
smarter than he was.
9.
Step A. This is an addition,
Step B.
f 0 (p)
=
( 1p )0
+
1
p
is added to
5
p2
.
( p52 )0
Step C. In order to avoid the quotient rule, we can write f (p) = p−1 + 5p−2 , and then we can use the power rule.
Step D. f 0 (x) = −p−2 − 10p−3
Step F. (optional) Diagram the function.
p−1
11.
+
5
p−2
Step A. This is a composite function. The outside is ( ).7 , and the inside is
−.3 1+v 0
Step B. f 0 (v) = .7( 1+v
) 13v
13v
Step C. This is a quotient, with
top 1 + v and bottom 13v.
−.3 13v(1+v)0 −(1+v)(13v)0 Step D. f 0 (v) = .7( 1+v
)
13v
(13v)2
−.3 −13 )
Step E. f 0 (v) = .7( 1+v
13v
(13v)2
1+v
13v
Step F. (optional) Diagram the function.
.7
1+v
13v
13.
Step
Step
Step
Step
Step
A. This is a composite function. The outside is tan( ) and the inside is z −
B. f 0 (z) = (sec2 (z − z1 ))(z − z1 )0
C. z − z1 is a difference.
D. f 0 (z) = sec2 (z − z1 )(1 + z12 )
F. (optional) Diagram the function.
1
.
z
Appendix E
537
tan
15.
Step
Step
Step
Step
Step
Step
z−
1
z
1
A. This is a composite function. The outside is ( ) 5 , and the inside is (j − 34)9 .
B. f 0 (j) = 51 ((j − 34)9 )−4/5 ((j − 34)9 )0
C. (j − 34)9 is a composite function, The outside is ( )9 , and the inside is j − 34. .
D. f 0 (j) = 15 ((j − 34)9 )−4/5 9(j − 34)8 (j − 34)0
E. f 0 (j) = 51 ((j − 34)9 )−4/5 9(j − 34)8 (1)
F. (optional) Diagram the function.
1
5
9
j − 34
Note: This problem more easily done by first computing
17.
Step
Step
Step
Step
Step
Step
(j − 34)9 = (j − 34)9/5 .
A. This is a product. z times sin(z) cos(z) (or you could arrange it as z sin(z) times cos(z)).
B. f 0 (z) = z 0 (sin z)(cos z) + z(sin z) cos z)0
C. sin(z) cos(z) is a product.
D. f 0 (z) = sin z cos z + z((sin(z))0 cos(z) + sin(z)(cos(z))0 )
E. f 0 (z) = sin z cos z + z cos2 z − z sin2 z
F. (optional) Diagram the function.
z
19.
p
5
Step A. This is a power, t to the
·
sin z
·
cos z
1
.
1000000
−.999999
Step B. f 0 (t) = t1,000,000
Step C. There aren’t any parts left to differentiate.
Step F. (optional) Diagram the function.
1
1,000,000
t
538
Answers, Notes, and Hints
'
In this chapter, we’re putting results before reasons, that
is, we are learning facts we will justify later, in Chapter 5.
This is how calculus progressed historically. The theorems
of the 18th century were only proved conclusively in the
19th. When challenged, the pioneers of calculus had different responses. Cavalieri dismissed critics, saying “Rigor is
the concern of philosophy and not of geometry.”
&
21.
$
%
Step A. This is a quotient. sin(x − 3) divided by x2 .
Step B. f 0 (x) =
x2 (sin(x−3))0 −(sin(x−3))(x2 )0
(x2 )2
Step C. The part sin(x − 3) requires the chain rule. The outside is sin() and the inside is x − 3. The part x 2 requires the
power rule.
Step D. f 0 (x) =
x2 (x−3)0 cos(x−3)−sin(x−3)2x
x4
x2 (cos(x−3))−sin(x−3)(2x)
x cos(x−3)−2 sin(x−3)
=
x4
x3
Step E. f 0 (x) =
Step F. (optional) Diagram the function.
sin
x−3
x2
23.
Step A. This is a quotient. (x sec(8x)) divided by (x + 3).
0
0
−(x sec(8x))(x+3)
Step B. f 0 (x) = (x+3)(x sec(8x))(x+3)
Step C. (x sec(8x)) is a product, x times sec(8x).
Step D. f 0 (x) =
(x+3)(x0 sec(8x)+x sec(8x)0 )−(x sec(8x)·1)
x2 +6x+9
(x+3)(sec(8x)+8x sec(8x) tan(8x))−x sec(8x)
x2 +6x+9
Step E. f 0 (x) =
Step F. (optional) Diagram the function.
x
·
sec
8x
x+3
25.
Step A. This is a sum, 7x2 is added to 3(12 − 5x17 )8 .
Step B. (7x2 )0 + (3(12 − 5x17 )8 )0
Step C. (12 − 5x17 )8 is a composite function.
Appendix E
539
Step D. f 0 (x) = 14x + 3(8(12 − 5x17 )7 ) · (12 − 5x17 )0 .
Step E. f 0 (x) = 14x + 24(12 − 5x17 )7 (−85x16 )
Step F. (optional) Diagram the function.
8
x2
7
27.
Step
Step
Step
Step
Step
Step
+
12 − 5x17
3
A. This is a product. (2x + 5) is multiplied by tan .3 (x2 )
B. f 0 (x) = (2x + 5)0 (tan(x2 )).3 + (2x + 5)(tan .3 (x2 ))0
C. The part (tan(x2 )).3 is a composite function. The outside function is ( ).3 . The inside is tan(x2 ).
D. f 0 (x) = 2(tan(x2 )).3 + (2x + 5).3(tan(x2 ))−.7 (tan(x2 ))0
E. f 0 (x) = 2(tan(x2 )).3 + (2x + 5).6x(tan(x2 ))−.7 (sec2 (x2 ))
F. (optional) Diagram the function.
.3
2x +
5
·
tan
x2
#
To those who compared integrals and derivatives to the
universally admired works of Greek mathematics, Blaise
Pascal expressed confidence that someday the methods of
Archimedes would justify calculus.
"
29.
Step A. This is a quotient. cot((12x + 5)3 ) is divided by tan(x −
Step B. f 0 (x) =
1 )(cot((12x+5)3 ))0 −tan(x− 1 )0 (cot((12x+5)3 ))
tan(x− x
x
1)
tan2 (x− x
!
1
).
x
Step C. tan(x − x1 ) is a composite fuction. The outside is tan( ) and the inside is x −
composite function. The outside is cot( ), and the inside is (12x + 5)3 .
Step D. f 0 (x) =
Step E. f 0 (x) =
1 )(− csc2 ((12x+5)3 )((12x+3)3 )0 )−sec2 (x− 1 )(x− 1 )0 (cot((12x+5)3 ))
tan(x− x
x
x
1)
tan2 (x− x
1 )(− csc2 ((12x+5)3 )(3(12x+5)2 )12)−(sec2 (x− 1 ))(1+ 1 ) cot((12x+5)3 )
tan(x− x
2
x
Step F. (optional) Diagram the function.
1)
tan2 (x− x
x
1
.
x
(cot((12x + 5)3 )) is also a
540
Answers, Notes, and Hints
3
cot
12x
tan
31.
Step
Step
Step
Step
Step
Step
−
5
1
x
A. This is a product. (x + 2) is multiplied by (3 csc(x + 2)) .
B. f 0 (x) = (x + 2)0 (3 csc(x + 2)) + (x + 2)(3 csc(x + 2))0
C. The part csc(x + 2) is a composite function, with the outside csc( ) and inside (x + 2).
D. f 0 (x) = 1(3 csc(x + 2)) + (x + 2)3(− csc(x + 2) cot(x + 2))(x + 2)0
E. f 0 (x) = (3 csc(x + 2)) − (3x + 6) csc(x + 2) cot(x + 2)
F. (optional) Diagram the function.
x
33.
x
+
+
2
·
3
csc
x+2
√
Step A. This is a quotient. (4x + 12)(sin(x + 3x2 )) is being divided by 2x2 x
√
√
x((4x+12)(sin(x+3x2 )))0 −(4x+12)0 (sin(x+3x2 ))(2x2 x)
√
(2x2 x)2
(4x + 12)(sin(x + 3x2 )) is a product, (4x + 12) times sin(x + 3x2 )
√
0
2
2 0
2√
)) )−(4x+12)(sin(x+3x2 ))(2x2 x)0
√
f 0 (x) = 2x x((4x+12) (sin(x+3x ))+(4x+12)(sin(x+3x
(2x2 x)2
√
2x2 x(4 sin(x+3x2 )+(4x+12) cos(x+3x2 )(1+6x))−(4x+12)(sin(x+3x2 ))(5x3/2 )
f 0 (x) =
4x5
Step B. f 0 (x) =
Step C.
Step D.
2x2
Step E.
Step F. (optional) Diagram the function.
Appendix E
541
4x +
12
·
2x2
35.
Step
Step
Step
Step
Step
Step
x + 3x2
sin
·
√
x
A. This is a product, 3x times (cot(3 + 4x3 ))9 .
B. f 0 (x) = (3x)0 (cot(3 + 4x3 ))9 + 3x((cot(3 + 4x3 ))9 )0
C. There is a composite function. The outside is ( )9 , and the inside is cot(3 + 4x3 ).
D. f 0 (x) = 3(cot(3 + 4x3 ))9 + 3x(9) cot8 (3x + 4)(cot(3x + 4x3 ))0
E. f 0 (x) = 3(cot(3 + 4x3 ))9 + 3x(9) cot8 (3 + 4x3 )(− csc2 (3 + 4x3 ))(12x2 )
F. (optional) Diagram the function.
9
3x
·
cot
3 + 4x3
'
$
&
%
The charge that somehow modern mathematicians were inferior to the ancients bothered Descartes. At one time, he
even wondered publicly whether the Greeks had, perhaps,
some methods for solving problems which they kept secret.
Leibniz voiced similar thoughts.
37.
Step A. This is a quotient. (15x4 )(x2 − 12) is divided by csc(2x + 4).
Step B. f 0 (x) =
Step C. The top
Step D. f 0 (x) =
csc(2x+4)((15x4 )(x2 −12))0 −(15x4 )(x2 −12)(csc(2x+4))0
csc2 (2x+4)
of the fraction is a product, (15x4 ) times (x2 − 12),
and csc(2x + 4) is a composite function.
csc(2x+4)((15x4 )0 (x2 −12)+(15x4 )(x2 −12)0 )−(15x4 )(x2 −12)(− csc(2x+4) cot(2x+4))(2x+4)0
(csc(2x+4))2
csc(2x+4)(60x3 (x2 −12)+(15x4 )(2x))−15x4 (x2 −12)(− csc(2x+4) cot(2x+4))(2)
csc2 (2x+4)
Step E. f 0 (x) =
Step F. (optional) Diagram the function.
542
Answers, Notes, and Hints
15x4
x2 − 12
csc 2x + 4
39.
Step A. This is a quotient. x5 + cot(x − 5x2 ) is divided by tan(sin(x)).
tan(sin(x))(x5 +cot(x−5x2 ))0 −(x5 +cot(x−5x2 )) tan(sin(x))0
(tan2 (sin(x)))
is a sum (x5 + cot(x − 5x2 )) and a composite function
Step B. f 0 (x) =
Step C. There
Step D. f 0 (x) =
Step E. f 0 (x) =
with outside tan( ) and inside sin(x).
tan(sin(x))((x5 )0 +cot(x−5x2 )0 )−(x5 +cot(x−5x2 ))(sec2 (tan(x))(sin(x))0
(tan2 (sin(x)))
tan(sin(x))(5x4 +(− csc2 (x−5x2 ))(1−10x))−(x5 +cot(x−5x2 ))(sec2 (sin(x)) cos(x))
tan2 (sin(x))
Step F. (optional) Diagram the function.
x5
+
tan
41.
cot
x − 5x2
sin
x
√
Step A. This is a quotient. (x + 5)( 12x + 5) is divided by (15x − 12x2 ).
Step B. f 0 (x) =
(15x−12x2 )((x+5)(
√
√
12x+5))0 −(x+5)( 12x+5)(15x−12x2 )0
(15x−12x2 )2
√
Step C. (x + 5)( 12x + 5) is a product.
Step D. f 0 (x) =
√
√
√
12x+5+(x+5)( 12x+5)0 )−(x+5)( 12x+5)(15−24x)
.
(15x−12x2 )2
√
√
)−(x+5)( 12x+5)(15−24x)
(15x−12x2 )( 12x+5+(x+5) √ 12
(15x−12x2 )((x+5)0
2 12x+5
Step E. f 0 (x) =
(15x−12x2 )2
Step F. (optional) Diagram the function.
Appendix E
543
x+5
15x
43.
√
12x + 5
− 12x2
Step A. This is a quotient. (5 − (sec(4x))3 ) is divided by x + 2x3 .
(x+2x3 )(5−(sec(4x))3 )0 −(5−(sec(4x))3 )(x+2x3 )0
(x+2x3 )2
5(sec(4x))3 is a difference.
Step B. f 0 (x) =
Step C.
Step D. f 0 (x) =
Step E. f 0 (x) =
(x+2x3 )(50 −((sec(4x))3 )0 )−(5−(sec(4x))3 )(1+6x2 )
(x+2x3 )2
(x+2x3 )(−3 sec2 (4x) sec(4x) tan(4x)(4))−(5−(sec(4x))3 )(1+6x2 )
(x+2x3 )2
Step F. (optional) Diagram the function.
3
5
−
x
sec
+
4x
2x3
'
$
&
%
Ultimately, according to Leibniz, we must move ahead
without complete justification. “It is not easy to demonstrate all the axioms, and to reduce demonstrations wholly
to intuitive knowledge. And if we had chosen to wait for
that, perhaps we should not yet have the science of geometry.”
45.
Step
Step
Step
Step
A. This is a composite function. The outside is ( )7 and the inside is 2x tan(x2 + 13)4 .
B. f 0 (x) = 7(2x tan(x2 + 13)4 )6 (2x tan(x2 + 13)4 )0 .
C. 2x tan(x2 + 13)4 is a product.
D. f 0 (x) = 7(2x tan(x2 + 13)4 )6 ((2x)0 tan(x2 + 13)4 + (2x)(tan(x2 + 13)4 )0 )
544
Answers, Notes, and Hints
Step E. f 0 (x) = 7(2x(tan(x2 + 13))4 )6 (2(tan(x2 + 13))4 + 2x(4 tan(x2 + 13)3 sec2 (x2 + 13)(2x)))
Step F. (optional) Diagram the function.
7
2x tan x2 + 13
31.
R4
33.
R1
35.
Integrating Elementary
Functions
Answers to exercises
from 4.7, page 214
37.
39.
1.
3x + C
3.
x2 + C
5.
πx + C
7.
9.
18
8x
9
41.
11.
2x2 + C
13.
xπ
15.
x5/2
17.
r2
19.
sin(w) + C
21.
3 sin(p) + C
23.
− cot(b) + C
25.
− csc(h) + C
27.
R1
29.
R1
0
0
4
3x2 dx = x3 = 64 − 1 = 63
1
−1
R2
1
1
4x3 dx = x4 2
2.6 dx = 2.6x = 5.2 − 2.6 = 2.6
1
−3
R3
2
45.
49.
51.
1
3 dx = 3x = 3 − 0 = 3
0
2x dx =
x2
1
=1−0=1
0
53.
2
2xπ =
1
2 · 2π − 2 · 1π = 2(2π − 1) ≈ 15.65
=
3
4 dt = 4t = 4(3) − 4(1) = 8
1
−1
1
9s8 dx = s9 −1
R .75
−.75
R
−3
(−1)−1 − (−3)−1 = − 32
=
2πxπ−1 dx
9x8 dx
+C
+C
−1
x− 1 =
2
1
R1
−x−2 dx
3
9x2 dx = 3x3 = 81 − 24 = 57
R2
1
= 1−1 = 0
−1
R −1
R3
47.
+C
1
43.
+C
x−3 + C
4
π
2
−π
2
= (1)9 − (1)9 = 2
.75
x9 =
−.75
(.75)9 − (−.75)9 ≈ .150
=
cos(x) dx
=
=
=
R .5
.25
R3
2
− csc2 x dx
=
=
− csc(x) cot(x) dx
π
2
sin(x) π
−2
sin( π2 ) − (sin(− π2 ))
1 − (−1) = 2
.5
cot(x) .25
cot(.5) − cot(.25) ≈ −2.086
=
=
3
csc(x) 2
csc(3) − csc(2) ≈ 5.986
Appendix E
'
On the subject of guessing to find integrals (or anything
else):
The man who makes no mistakes does not usually make
anything.
Edward John Phelps
&
545
'
$
Think of the Fundamental Theorem as linking two very different processes. One, Euler’s method, the limit of sums,
and two, the difference quotient, the limit of ratios. The
discovery that one was the inverse of the other was the key
that unlocked everything.
&
%
19.
R4√
y + 1 dy
0
=
=
Integrating Complicated
Functions
=
Answers to exercises
from 4.8, page 217
1.
R3
3.
R1
1
5.
7.
R1
9.
11.
13.
15.
17.
1
0
23.
3
4 dt = 4t = 4 · 3 − 4 · 1 = 12 − 4 = 8
1
−1
R3
21.
1
2r dr = r 2 25.
1
1
w 3 dw = 41 w 4 =
2
R1
− 14 04 =
=
=
cos(t/2) dt
R8 √
3
x dx
1
=
=
=
=
4
3 2 2
x
3
3
2
(4) 2
3
14
3
=
=
=
=
=
(x3 + 2) dx
R −1
=
(2)4
4
=
=
=
1
2
2t2 − 3t + 5 dt
=
=
1
sec(1 + z) R3
2
h − 1 dh
−1
sec(2) − sec(0)
sec(2) − 1
29.
R − 14
5
−4
=
=
−4m − 1 dm
=
=
=
31.
2 sin(1) − 2 sin(0) = 2 sin(1)
1
4
4
3
(8) 3 − 43 (1) 3
4
3
1
12 − 4 = 11 4
33.
2
R0
− 3r4 dr
R3
(3p − 1) dp
−2
1
4
2 3
t − 23 t2 + 5t 3
2
3
2
3
2
(4) − 2 (4) + 5(4) −
3
2
(2)3 − 32 (2)2 + 5(2)
3
− 48
+ 20)−
( 128
3
2
16
( 3 − 12
+ 10)
2
88
3
1 2
h −
2
1
(3)2
2
3
2
=
3
0
8
−3
6(−1)−1 − 6(−3)−1
−4
=
27.
=
=
=
=
=
=
0
+ 2(2) − 0 = 8
−1
6a−1 =
− 32 (1) 2
2
2 sin( 2t ) 4
3 3
x
4
R4
− 1)
2
+ 2x x4
4
−6a−2 da
−3
2 2/3
(5
3
=
1
4
3
z + 21 z 2 2
3 + 12 32 − 2 + 21 22 = 3 12
sec(1 + z) tan(1 + z) dz
R4√
x dx
1
0
=
=
−1
R2
1 4
1
4
0
(1 + z) dz
0
=1−1=0
−1
3
3x2 dx = x3 = 33 − 13 = 27 − 1 = 26
R3
R2
4
3 + 1) 2 0
3
2
(4 + 1) 2 −
3
3
2
(0 + 1) 2
3
2
(y
3
3
h
2 − (3) −
1
(2)2
2
− (2)
− 1
4
−2m2 − m 5
−4
−2(− 14 )2 − (− 14 ) −
5 2
5
−2(− 4 ) − (− 4 )
1
− 15
= − 74
8
8
0
− 41 r 3 −2
− 14 (0)3 − − 41 (−2)3
0 − (−2) = 2
3
3 2
p −p
2
1 3
(3)2 − (3)
2
1
21
− 2 = 10
2
−
3
(1)2
2
− (1)
$
%
546
35.
37.
39.
Answers, Notes, and Hints
Rπ
0
R3
2
− sin(x) dx
4
x 3 dx
R 1.5
1 4
−1.5 3 x
(cos(π)) − (cos(0))
(−1) − (1) = −2
3
=
7
3 3
x
7
=
7
3
(3) 3
7
dx
2
7
− 73 (2) 3
1.5
1 5
x 15
=
−1.5
1
1
5
(1.5)5 − 15
15 (−1.5)
1
35
35
+ 25
3·5 25
81
80
=
=
43.
0
=
=
=
41.
π
cos(x) =
You may have guessed sin(3x), which is close. However,
(sin(3x))0 = 3 cos(3x). Notice that we need a coefficient
of 6 in front of cos(3x). So if we begin with 2 sin(3x), and
differentiate, we will get 6 cos(3x). Now, we can compute
π
R
the definite integral, 0π 6 cos(3x) dx = 2 sin(3x) =
R
cos(s) ds − 13 s23 ds
3 R
sin(s) − 13 2s−3 ds
1
3
sin(3) − sin(1) − −s−2 R
8f − 5 df = 4f 2 − 5f + C
61.
R
4j 1.2 dj =
63.
R
3(y − 2)4 dy =
65.
R
7 cos(Θ) dΘ = 7 sin(Θ) + C
67.
R
1
n2
69.
R
sin(m) +
1
m3
71.
R
sin(x)
cos2 (x)
=
=
73.
0
2 sin(3π) − 2 sin(0) = 0.
R3
cos(s) − s23 ds =
1
59.
77.
1
sin(3) − sin(1) − − 91 − −1
sin(3) − sin(1) − 98 .
=
=
45.
Skip this problem! The function is not defined for all
points in the interval −1 to 1.
47.
Consider that in this problem, the coefficient of sin(c)
is the derivative of the argument of sin(c). This tells
us that the function we are looking for would probably need the product
rule in order to differentiate it.
R
We know that (− sin(c)) = cos(c). So, by combining these two ideas, our guess for the antiderivative
might be cos(3c2 + 2c − 1). Let’s check: (cos(3c2 +
2c − 1))0 = − sin(3c2 + 2c − 1)(3c2 + 2c − 1)0 =
−(6c + 2) sin(3c2 + 2c − 1). Since this is what we were
looking for, we can now compute the definite integral.
0
R0
−(6c+2) sin(3c2 +2c−1) = cos(3c2 + 2c − 1) =
−1
z(2) − 1
z(2)
79.
(cos(3(0)2 + 2(0) − 1)) − (cos(3(−1)2 + 2(−1) − 1)) =
cos(−1) − cos(0) = cos(−1) − 1.
R
sin(7y) dy = − 71 cos(7y) + C
53.
R
55.
R
57.
(z 3 − 3z + 2z 2 ) dz =
1 4
z
4
− 2)5 + C
n−2 − n4 dn
−n−1 − 15 n5 + C
R
dm = − cos(m) − 21 m−2 + C
R
R
sin(x)
1
cos(x) cos(x)
sec(x) tan(x) = sec(x) + C
=
=
=
=
=
=
=
=
R2
sin(t) dt
2
cos(t) 0
− cos(2) + cos(0)
− cos(2) + 1
2 − cos(2)
0
R5
1
dt
3 t2
5
−1 t
3
− 51 + 13
17
15
Note: The problemRshould have read: s0 = t2 − t.
3 2
s(3) − s(1) =
1 (t − t)dt
3
3
t2 t
−
=
3
2 s(3) − 3
s(3)
−1
51.
y(5) − y(3)
y(5) − 1
y(5)
83.
3
(y
5
=
=
=
81.
+C
2x sec2 (x2 ) = tan(x2 ) + C
1
=
R
− n4 dn
z(2) − z(0)
R3
=
49.
R
4 2.2
j
2.2
=
=
=
− 13 −
− 13 −
2 16
1
2
1
2
1
−3
Note: The problem should have read: x0 = (t + 2)2 .
R3
x(3) − x(1) =
(t + 2)2 dt
1
3
(t+2)3 =
3
−x(1)
x(1)
− 32 z 2 + 23 z 3 + C
=
=
=
53
3
−
32 23
−32 23
1
2
33
3
1
csc(4s) cot(4s) ds = − 41 csc(4s)
85.
R
sin(3 − 2p) dp =
1√
w
w
87.
R
t sin(t) dt = sin(t) − t cos(t) + C
dw
=
=
R
R
1
w −1 w 2 dw
1
w− 2
R √
R 3√
t 5t dt = t 2 5 dt =
dw =
1
2w 2
√
2 5 5
t2
5
+C
+C
89.
R √
x x + 1 dx =
2
(x
5
cos(3 − 2p) + C
+ 1)5/2 − 32 (x + 1)3/2 + C
Appendix E
547
21.
Exponential Growth
and Decay
23.
Answers to exercises
from 4.9, page 227
25.
1.
3.
y0
y(0)
y(x)
y(3)
=
=
=
=
=
2e9
Ae 5 x
=
7
Ae 5 (2)
=
=
(7x−14)
5
e
21
e5
=
=
=
1
1
14
e− 5
g(x)
g(5)
=
=
14
e− 5
14
e− 5
d0
=
d( 53 )
=
7
A
=
=
7
5
7e 2
d(x)
d(1)
=
=
5
7e 2
5
7e 2
m0
=
πm, m(x)
=
Aeπx
m( π2 )
=
−1
=
Aeπ( 2 ) = Ae
=
−1
A
=
m(x)
=
m(π)
g(x)
·
·
−3d
,
2
−e
7
e5x
7
e 5 (5)
d(x)
1 r/2
e
2
√
1
3
f (p) = e3p+1 = (e3p+1 ) 3 = ep+1/3
0
p+1/3
f (p) = e
(p + 1/3)0 = ep+1/3
p0 (x) =
√
e√x
2 x
Who has not been amazed to learn that the
function ex , like a phoenix rising again from
its own ashes, is its own derivative?
14
5
– Francois le Lionnais
&
Ae− 2 x
=
3 5
Ae− 2 ( 3 )
5
= Ae− 2
The Natural Logarithm
·
·
=
=
(5−3x)
7e 2
7e
π
π2
2
2
− π2
π2
2
π2
−e− 2
· eπx
, j(x)
− 2j
3
π
−e 2 (2x−π)
=
· eπ(π)
π2
−e 2
=
2
=
Ae− 3 x
=
Ae− 3 (0) = Ae0 = A
=
5e−
j0
=
j(0)
j(x)
j(7)
=
=
=
5
2
5 · e− 3 x
−2
5 · e 3 (7)
q0
q(0)
q(x)
q(7)
=
=
=
=
.14q, q(x)
−5 = Ae.14(0)
−5e.14x
−5e.14(7)
1.
f 0 (s) = es cos(es )
3.
f 0 (d) = e−2d (−2d2 )0 = −4de−2d
5.
f 0 (g) =
7.
f 0 (h)
2
14
3
=
=
Ae.14x
Ae0 = A
=
−5e.98
15.
h0 (y) = sin(y)e3+y + cos(y)e3+y
17.
w 0 (z) = ee ez
2
2g+4
ln(2)(g 2 +4g)
(e−h )0 (4h2 )−(4h2 )0 (e−h )
(4h2 )2
−4h2 e−h −8he−h
2
(4h )2
=
=
2
f 0 (x) = ex−7
1
4+x
9.
f 0 (x) =
11.
f 0 (r) = er 2r
13.
f 0 (z)
=
=
(cos2 (z))(ln(z))0 + (ln(z))(cos 2 (z))0
(cos2 (z))( 1z ) + (ln(z))(−2 sin(z) cos(z))
15.
f 0 (y)
=
=
=
1
(cos(4y))0 − (2y)0
cos(4y)
1
(− sin(4y))(4) − 2
cos(4y)
2
z
r 0 (t) = (e4t )0
= e4t (4t)0 = 4e4t
=
3
=
13.
19.
er
√
er
Answers to exercises
from 4.10, page 235
3
e− 2 x
3
e− 2 (1)
−e−
=
= Ae
2
'
7
=
g(2)
14
Ae 5
A
π2
Ae 2
11.
Ae3x
Ae3(0) = A · 1 = A
=
5
Ae− 2
9.
7g
,
5
=
=
g0
5.
7.
3y, y(x)
2
2 · e3x
2 · e3(3)
s0 (r) =
17.
f (x)0 =
−4 tan(4y) − 2
sec2 (x)
ln(3) tan(x)
$
%
548
Answers, Notes, and Hints
f 0 (s)
19.
1
(e2s + sin(es ))0
e2s +sin(es ) 1
(e2s )(2) + cos(es )(es )
e2s +sin(es )
2s
s
s
2e +e cos(e )
e2s +sin(es )
=
=
=
21.
f 0 (z)
25.
R
27.
R
29.
R
(sin(e2z ))0 cos(e2z ) + (cos(e2z ))0 sin(e2z )
2e2z cos2 (e2z ) − 2e2z sin2 (e2z )
=
=
1
x+2
dx = ln |x + 2| + C
2x
x2 −3
Clearing Marian Blake
dx = ln |x2 − 3| + C
section 4.10, p. 242
e3−2y dy = − 21 e(3−2y) + C
Euler was the first to use the symbol “e” for the number
2.71828. . . .
R
31.
e3v
dv =
1 3v
e
3
33.
35.
R2
R0
5 sec2 (5x)etan(5x) dx
R
39.
R
41.
p
1 + e3h dh
2
e−πn n dn
ax dx =
(logx (2))0
43.
=
1
ax
ln(a)
=
The equation is (TO − TR )0 = −k(TO − TR ), where TR is the
temperature of the room (constant) and TO is the temperature of the object. If we let d = TO − TR , then d0 = −kd, so
d = Ae−kt .
2
etan(5x) =
0
etan(10) − etan(0)
etan(10)−1
=
=
3h
−2 e
37.
Note that the body will remain at 98.6 (despite the chilly
vault) until the point of death.
+C
R√
1
et dt = 2e 2 t + C
0
Here’s another question for you: The halflife of an element is
the length of time it takes for the element lose half its mass.
Given the value of k (.0001245), what is the halflife of carbon
14?
Dating Languages
Answers to exercises
from 4.11, page 246
0
2
3h ) 3
2 (1
+
e
9
=
2 1.5
(2
9
=
−2
− (1 + e−6 )1.5 )
2
1 −πn
+C
e
− 2π
+ C.
From
3.
From 124 = Ae2ktnow = 210e2(−.000217)tnow we
= e2(−.000217)tnow , then taking the log of
get 124
210
both sides, ln 124
= 2(−.000217)tnow , so tnow =
210
1
124
= 1213.88 . . .. This tells us that
ln 210
2(−.000217)
English split off from German about 1200 years ago, or
around the year 800.
5.
1
Computing tnow = ln 144
, we get
210 −.0002291137
1646.85 . . ., a change of about 90 years and enough to
put us in the interval mentioned by Professor Gregory.
ln(2) 0
( ln(x)
)
ln(2)
=
=
47.
ln(x)·0− x
(ln(x))2
ln(2)
− x(ln(x))
2
(xx )0 = (ex ln(x) )0
45.
R
=
=
=
ex ln(x) (x ln(x))0
ex ln(x) (ln(x) + x x1 )
xx (ln(x) + 1).
tan(x) dx = − ln | cos(x)| + C.
Carbon Dating
section 4.10, p. 238
144 = N (tnow ) = 210ektnow we get
= e−.000217tnow , then taking the log of
both sides, ln 144
= −.000217tnow , so tnow =
210
1
144
= 1738.68 . . .. This gives us a date
ln 210
−.000217
around 250, not in the range suggested by Professor Gregory.
1.
144
210
Appendix E
Applications of mathematics like this are controversial. Does
this model contribute to our understanding of the evolution
of languages? We’re not glottochronologists ourselves, so our
opinion is neither definitive or significant. Nonetheless, here’s
our opinion: We think this is fun.
549
It is clear that Economics, if it is to be a science at all, must be a mathematical science.
– William Jevons
Elasticity
Answers to exercises
from 4.12, page 251
Gravity
Answers to exercises
from 4.13, page 256
1.
x
3
3x+17
3.
w
4w6
5.
x
3e4x
7.
z
3z−ez
9.
v
5vev
11.
13.
12e4x = 4x
(3 −
3.
y
y2
3y+5
2y(3y+5)−3y 2
(3y+5)2
ln(y)0
ln(x)0
=
3y+10
3y+5
x dy
y dx
Let w = f (x)g(x), y = f (x), z = g(x). Then
=
=
x
yz
x 0
w
w
zy
yzzx
+ xyx
x
=
=
5.
p 0 dR
n , dp
n
=
+ zy 0 )
zx + yx .
−15
−9.8
=
=
=
1.5306
h(1.5306)
x 0 0
(z y )
z
7.
= n + pn0 , then
dR
dp
a(t)
v(t)
h(t)
v(t)
0
x
(yz 0
yz
Let z = g(f (x)), y = f (x). Then
np =
−100
−4.9
4.5175
v(4.5175)
zx
19.
0
q
1 (y 0 )
y
1
x
=
=
=
=
=
a(t)
v(t)
h(t)
(5ev + 5vev ) = 1 + v
wx
17.
a(t)
v(t)
−20
−20
−9.8
ez )
=
15.
1.
24w 5 = 6
a(t)
v(t)
0
n
n + p pnp
n + nnp
n(1 + np ).
21.
If np < −1, then R0 < 0, so increasing price results in
decreasing revenue.
23.
If np = −1, then R0 = 0, so increasing price results
have do not change the revenue.
=
=
=
=
=
h(t)
−15
h( −9.8
)
9.
a(t)
v(t)
v(6.3004)
11.
y
−9.8
−4.9t2
−4.9t2 + 100
0
−4.9t2 + 100
=
=
=
≈
t
t
−9.8(4.5175)
−44.2715 m/sec.
=
=
=
=
=
=
=
≈
−9.8
−9.8t + 15
−4.9t2 + 15t
0
−9.8t + 15
t
t
11.48 m.
=
=
=
=
=
≈
−15
−9.8
=
=
−9.8
−9.8t
−9.8t
t sec.
=
=
≈
−9.8
−9.8t + 15
−9.8t + 15
t
−4.9t2 + 15t + 100
111.48 m.
−9.8
−9.8t + 15
−46.744 m/sec.
15 √
cos(45)
15 22 ≈ 10.6
550
Answers, Notes, and Hints
13.
h(
a(t)
v(t)
0
=
=
=
t
h(t)
=
=
15
√ )
9.8 2
=
15.
a(t)
v(t)
d(t)
d(15
17.
1√
)
4.9 2
ay (t)
vy (t)
h(t)
0
z sin(α)
4.9
ax (t)
vx (t)
d(t)
=
=
=
=
=
=
=
=
=
−9.8
−9.8t + 15 cos( π4 )
−9.8t + 15 √1
15
√
9.8 2
Assignment #1:
2
sec.
−4.9t2 + 15 √1 t
1.
≈ 5.74 m.
3.
152
9.8·4
=
=
=
2
0
15 cos( π4 )
15
√
t
2
2
= 15
9.8
≈ 22.96 m.
−9.8
−9.8t + z sin(α)
−9.8t + z sin(α)t
−9.8t + z sin(α)t
t
0
z cos(α)
z cos(α)t
z 2 sin(α) cos(α)
4.9
−9.8
≈ −29.873 m/s.
−9.8
≈ −4136.270 m/s.
6π(1.81×10− 5)(.0005)
.52×10− 6
6π(1.81×10− 5)(.005)
7.2×10− 4
Assignment #2:
1.
kt gets bigger. In the limit,
w = eA
kt . As t gets bigger, e
w goes to zero.
3.
Again, w goes to zero and v goes to − 9.8
.
k
Trigonometric Functions
m.
Answers to exercises
from 4.15, page 266
19.
$
'
Object 1:
a(t)
v(t)
h(t)
=
=
=
−9.8
−9.8t
−4.9t2 + 200 m.
Object 2:
ay (t)
vy (t)
h(t)
ax (t)
vx (t)
d(t)
100
100
sec.
z cos(α)
=
=
=
=
=
=
=
=
−9.8
−9.8t + z sin(α)
−4.9t2 + z sin(α)t
0
z cos(α)
z cos(α)t
z cos(α)t
t
200
tan(α)
α
=
=
=
100
z sin(α) z cos(α)
2
arctan(2)
Trigonometry, so far as this, is most valuable to every man. There is scarcely a day in
which he will not resort to it for some of the
purposes of common life.
&
Answers to exercises
from 4.14, page 259
%
1.
90 degrees is π2 radians. 180 degrees is π radians. In genπx
eral, 180x degrees is πx radians, or x degrees is 180
ra◦
◦
πx
πx
dians. Thus, sin(x) = sin 180 and cos(x) = cos 180 .
3.
sin(x)0
=
sin
π
πx
( 180 ).
cos 180
5.
cos(x)0
◦
◦
− sin
7.
Terminal Velocity
– Thomas Jefferson
◦
=
πx
180
=
0
=
0
πx
=
180
◦
π
− 180 sin(x).
cos
π
180
πx
180
cos
πx
180
πx
180
− sin
πx
sec(x) = sec( 180
)
◦
πx 0
180
=
πx 0
180
=
◦
πx 0
πx
πx
π
π
) = sec( 180
) tan( 180
)( 180
) = sec(x)tan(x)( 180
)
sec( 180